Question

The circuit shown in the figure has ideal opamp, the circuit acts as

• A. band stop filter
• B. band pass filter
• C. low pass filter
• D. high pass filter
  

Answer 1

The correct option is D high pass filter


Let us solve the circuit at $\omega =0$
At $\omega =0$ the capacitors are open ckt

so ,

clearly $V_0=0$
so, gain at $\omega =0$ is zero
Now at $\omega =\infty$ the capacitors are short circuited so the circuit is



so, $V_0=\frac{-R_2}{R_1}{V}_i$

so, magnitude of gain $=\frac{R_2}{R_1}$
Thus at low frequencies gain = 0
and at high frequencies gain=$\frac{-R_2}{R_1}$

thus it is a HPF (High pass filter )